Tenaha is a town in Shelby County, Texas, United States. The population was 1,160 at the 2010 census.
Tenaha is located at 31°56′39″N 94°14′47″W / 31.94417°N 94.24639°W / 31.94417; -94.24639 (31.944062, -94.246274).
According to the United States Census Bureau, the town has a total area of 4.0 square miles (10.3 km²), of which 4.0 square miles (10.2 km²) is land and 0.04 square miles (0.1 km²) (0.50%) is water.
As of the census of 2010, there were 1,160 people, 422 households, and 287 families residing in the town. The population density was 268.4 people per square mile (103.6/km²). There were 475 housing units at an average density of 122.3 per square mile (47.2/km²). The racial and ethnic makeup of the town was 37.6% non-Hispanic African American, 36.6% non-Hispanic White, 24.0% Hispanic or Latino of any race, and 1.9% other.
Of the 422 households, 36.0% had children under the age of 18 living with them, 41.2% were married couples living together, 20.6% had a female householder with no husband present, and 32.0% were non-families. 26.3% of all households were made up of individuals living alone; 12.3% had someone living alone who was 65 years of age or older. The average household size was 2.75 and the average family size was 3.33.
In the town, the population was spread out with 32.2% under the age of 18, 8.7% from 18 to 24, 26.7% from 25 to 44, 20.7% from 45 to 64, and 11.8% who were 65 years of age or older. The median age was 32 years. For every 100 females, there were 85.5 males. For every 100 females age 18 and over, there were 80.4 males.
The median income for a household in the town was $23,750, and mean household income was $31,055. Median family income was $26,154, and mean family income was $32,600. The town’s per capita income was $11,600. About 33.6% of families and 35.9% of the population were below the poverty line, including 46.4% of those under age 18 and 30.4% of those aged 65 or over.